$H_2(G,\mathbb{Q})$ for finitely presented group $G$

Question

I am learning about group homology and spectral sequences and I have read somewhere that we can use the Lyndon-Serre-Hochschild spectral sequence to prove that $H_2(G,\mathbb{Q})$ has finite rank when $G$ is finitely presented but I cannot write it down.

I have tried applying the LSH spectral sequence to the exact sequence $1\to N\to F\to F/N\to 1$ with $F$ free and $G\cong F/N$ but I don't reach any conclussions.

Can anybody tell me how to do that?

Answer 1

After several misfires, I have it now.

Let $G$ be a finitely presented group; there is a free group $F$ on $n\in\Bbb N$ letters and a subset $R$ of $m$ words of $F$, and a specified isomorphism $G\cong F/\overline{\langle R\rangle}$ where the bar denotes normal closure and the braces denote subgroup generation.

The Lydon-Hochschild-Serre spectral sequence applied to the trivial $F$-module $\Bbb Q$ and the pair $(G,\overline{\langle R\rangle})$ is convergent to $H_\ast(F;\Bbb Q)$ (under some mysterious filtration) and has $E^2$ page: $$E^2_{p,q}\cong H_p(G;H_q(\overline{\langle R\rangle};\Bbb Q))\implies H_{p+q}(F;\Bbb Q)$$

Here, $H_q(\overline{\langle R\rangle};\Bbb Q)$ is abstractly - but canonically - a $F/\overline{\langle R\rangle}\cong G$-module. It will be very important to understand specifically what the $G$-action is. I investigated the generalities of this question here (thanks again to Mariano). Even though $\Bbb Q$ is a trivial module, this homology module is not necessarily a trivial $G$-module.

There is the $5$-term exact sequence in low degree associated to the spectral sequence: $$H_2(F;\Bbb Q)\to H_2(G;H_0(\overline{\langle R\rangle};\Bbb Q))\to H_0(G;H_1(\overline{\langle R\rangle};\Bbb Q))\to H_1(F;\Bbb Q)\\\to H_1(G;H_0(\overline{\langle R\rangle};\Bbb Q))\to0$$

We need to understand these terms. Firstly (see Weibel's "Homological Algebra" chapter $6$ as a reference for my claims) $H_2(F;\Bbb Q)=0$ since $F$ is a free group. $H_1(F;\Bbb Q)\cong F^{\mathsf{ab}}\otimes_{\Bbb Z}\Bbb Q\cong\Bbb Q^n$ where the first isomorphism comes from the fact $\Bbb Q$ is here considered a trivial $F$-module and the second isomorphism from the fact $F^{\mathsf{ab}}\cong\Bbb Z^n$. Similarly, by triviality of $\Bbb Q$ as an $\overline{\langle R\rangle}$-module, the $H_0$ terms are just isomorphic to $\Bbb Q$ and the $H_1$ term is isomorphic to $\overline{\langle R\rangle}^{\mathsf{ab}}\otimes_{\Bbb Z}\Bbb Q$.

So the sequence is: $$0\to H_2(G;\Bbb Q)\to H_0(G;\overline{\langle R\rangle}^{\mathsf{ab}}\otimes_{\Bbb Z}\Bbb Q)\to\Bbb Q^n\to H_1(G;\Bbb Q)\to0$$However, for the sake of rigour it is our duty to figure out how "$\Bbb Q$" has inherited a $G$-module structure in the first and final nonzero terms. We need it to be the trivial structure, but a priori $H_0(\overline{\langle R\rangle};\Bbb Q)$ inherits a mysterious $G$-action.

The isomorphism $H_0(\overline{\langle R\rangle};\Bbb Q)\cong\Bbb Q$ is induced from - if you compute $H_0$ via a bar resolution - $\Bbb Z[\overline{\langle R\rangle}]\otimes_{\Bbb Z[\overline{\langle R\rangle}]}\Bbb Q\cong\Bbb Z[\overline{\langle R\rangle}]\otimes_{\Bbb Z}\Bbb Q\to\Bbb Q$, the map induced by $(\sum_i r_in_i)\otimes q\mapsto \sum_i n_iq$. In particular, $q\in\Bbb Q$ is identifiable with $1\otimes q\in\Bbb Z[\overline{\langle R\rangle}]\otimes_{\Bbb Z[\overline{\langle R\rangle}]}\Bbb Q$. The $G$-action on the bar resolution (see my linked post, but dualise since we use a right bar resolution now) would map $q$ to $gq$ in the $g(1)g^{-1}$th summand i.e. map $1\otimes q$ to $1\otimes q\sim q$. Thus, the inherited $G$-action on $H_0(\overline{\langle R\rangle};\Bbb Q)\cong\Bbb Q$ is the trivial one and we are justified in writing "$H_2(G;\Bbb Q),H_1(G;\Bbb Q)$" in the exact sequence.

However, the action on $H_1(\overline{\langle R\rangle};\Bbb Q)$ usually is not trivial. Let $\pi:\overline{\langle R\rangle}\to\overline{\langle R\rangle}^{\mathsf{ab}}$ denote the Abelianisation map. It can be shown that $H_1(\overline{\langle R\rangle};\Bbb Q)\cong\overline{\langle R\rangle}^{\mathsf{ab}}\otimes_{\Bbb Z}\Bbb Q$ is induced from the following map on the bar resolution: $B_1\otimes_{\Bbb Z[\overline{\langle R\rangle}]}\Bbb Q\to\overline{\langle R\rangle}^{\mathsf{ab}}\otimes_{\Bbb Z}\Bbb Q$, $\left(\sum_i[r'_i]r_in_i\right)\otimes q\mapsto\left(\sum_in_i\pi(r'_i)\right)\otimes q$.

Let $x=\pi(r)\in\overline{\langle R\rangle}^{\mathsf{ab}}$. $x\otimes q$ is represented by $[r]\otimes q$ in the bar model. The $G$-action by some $g$ is, according to my post, going to be $g(x\otimes q):=[grg^{-1}]\otimes gq=[grg^{-1}]\otimes q$. Now here's the thing; every element of $\overline{\langle R\rangle}$ is obtainable, by conjugating finitely many times, from an element of $\langle R\rangle$. Zeroth homology takes coinvariants and identifies $\pi(grg^{-1})\otimes q\sim[grg^{-1}]\otimes q\sim x\otimes q$. It follows all zeroth homology classes will be representable by an element of $\pi(\langle R\rangle)\otimes_{\Bbb Z}\Bbb Q$.

However, $\pi(\langle R\rangle)\otimes_{\Bbb Z}\Bbb Q$ is an Abelian group generated by at most $m$ elements tensored with $\Bbb Q$ and thus is a $\Bbb Q$-vector space of dimension $d\le m$. When we take zeroth homology, we are going to take a quotient of this and this again results in a $\Bbb Q$-vector space of dimension $\le m$. Therefore the exact sequence is: $$0\to H_2(G;\Bbb Q)\to\Bbb Q^k\to\Bbb Q^n\to H_1(G;\Bbb Q)\to0$$Where $0\le k\le m$.

It follows that $H_2(G;\Bbb Q)$ is a $\Bbb Q$-vector space of dimension $d\le m$ and that $H_1(G;\Bbb Q)$ is a $\Bbb Q$-vector space of dimension $n-m\le d\le n$. This shows these groups are of finite rank and gives you an estimate on the size of this rank.

Answer 2

Here is an idea: Since $G$ is finitely presented, we can write $G \cong \langle F\rangle / \langle R \rangle$ with $|F|,|R| < \infty$. You probably want to use the five-term exact sequence which is a direct consequence of the LSH spectral sequence, see Corollary 6.4 on page 171 in Brown's book. Using the fact that $H_1(G;M) \cong G^{\text{ab}} \otimes M$ for $M$ an abelian group with trivial $G$-action, the first three terms of the five-term exact sequence applied to $1 \to \langle R\rangle \to \langle F\rangle \to G \to 1$ yield an exact sequence $$ H_2(F;\mathbb{Q}) \to H_2(G;\mathbb{Q}) \to \langle R\rangle^{\text{ab}} \otimes \mathbb{Q}. $$ Notice that $R^{\text{ab}} \otimes \mathbb{Q}$ is torsion-free and has finite rank (since $R$ was finite). Your claim should now follow, since $H_2(F;\mathbb{Q}) \cong 0$, i.e. the second map is injective.

Feel free to point out any mistakes here; I did not think everything through rigorously.